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          BZOJ 4513  [Sdoi2016]储能表
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<p>有一个 n 行 m 列的表格，行从 0 到 n−1 编号，列从 0 到 m−1 编号。每个格子都储存着能量。最初，第 i 行第 j 列的格子储存着 (i xor j) 点能量。所以，整个表格储存的总能量是，<span class="math display">\[\sum\limits_{i = 0} ^ {n - 1} \sum\limits_{j = 0} ^ {m - 1} (i \ {\rm xor} \ j)\]</span> 随着时间的推移，格子中的能量会渐渐减少。一个时间单位，每个格子中的能量都会减少 1。显然，一个格子的能量减少到 0 之后就不会再减少了。 也就是说，k 个时间单位后，整个表格储存的总能量是，<span class="math display">\[\sum\limits_{i = 0} ^ {n - 1} \sum\limits_{j = 0} ^ {m - 1} \max((i \ {\rm xor} \ j) - k, 0)\]</span> 给出一个表格，求 k 个时间单位后它储存的总能量。 由于总能量可能较大，输出时对 p 取模。</p>
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<p><a target="_blank" rel="noopener" href="http://www.lydsy.com/JudgeOnline/problem.php?id=4513">http://www.lydsy.com/JudgeOnline/problem.php?id=4513</a> 好难的数位DP啊...感觉网上并没有太好的题解，理解了好长时间... 我主要参考的题解：<a target="_blank" rel="noopener" href="http://fancypei.github.io/2016/04/16/SDOI2016%20Round1/#排列计数">http://fancypei.github.io/2016/04/16/SDOI2016%20Round1/#排列计数</a> 其实是数位DP的传统套路，按位进行讨论，讨论的时候，没有上限（下限）的时候可以枚举所有的0和1，但是之前的所有位到达上限（下限）的时候，就不能枚举了，只能和之前的一样，或者限制的值为1，可以枚举0（限制的值为0，可以枚举1）。 那么我们的状态设计就是： <span class="math inline">\(f[x][0/1][0/1][0/1]\)</span>表示从最高位枚举到了第<span class="math inline">\(x\)</span>位的全部有贡献的状态的结果和，第二维代表枚举值和<span class="math inline">\(n\)</span>的关系，第三维表示枚举值和<span class="math inline">\(m\)</span>的关系，第四维表示枚举值和<span class="math inline">\(k\)</span>的关系，值为0，就代表之前的位还没有到达限制，这一位可以随便枚举，值为1，代表之前的位已经到达限制了，这一位需要被原有值的限制。 <span class="math inline">\(g[x][0/1][0/1][0/1]\)</span>表示从最高位枚举到了第<span class="math inline">\(x\)</span>位的全部有贡献的状态数，状态设计和上一个数组同理。 然后枚举上一位的状态<span class="math inline">\(a,b,c\)</span>和这一位的数值<span class="math inline">\(xx,yy,zz\)</span>，通过给定的这一位数值，来计算出这一位的状态<span class="math inline">\(aa,bb,cc\)</span>。 对于方案数的状态转移非常简单，就是<span class="math display">\[g[i][aa][bb][cc]+=g[i+1]_{abc}\]</span> 对于结果的话，要计算出这一位对于结果的贡献，实际上，二进制的减法运算可以等效为列竖式的样子，各个位数的结果可以独立，到时候相加就可以了。 知道这个理论的话，方案的结果转移也比较简单<span class="math display">\[f[i][aa][bb][cc]+=(zz-z)\times 2^i \times g[i+1]_{abc}\]</span></p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span><span class="meta-string">&lt;bits/stdc++.h&gt;</span></span></span><br><span class="line"><span class="keyword">using</span> <span class="keyword">namespace</span> std;</span><br><span class="line"><span class="keyword">long</span> <span class="keyword">long</span> T,n,m,k,mod,f[<span class="number">62</span>][<span class="number">2</span>][<span class="number">2</span>][<span class="number">2</span>],g[<span class="number">62</span>][<span class="number">2</span>][<span class="number">2</span>][<span class="number">2</span>];</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span></span><br><span class="line"><span class="function"></span>&#123;</span><br><span class="line">    <span class="built_in">scanf</span>(<span class="string">&quot;%lld&quot;</span>,&amp;T);</span><br><span class="line">    <span class="keyword">while</span>(T--)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="built_in">memset</span>(f,<span class="number">0</span>,<span class="built_in"><span class="keyword">sizeof</span></span>(f));</span><br><span class="line">        <span class="built_in">memset</span>(g,<span class="number">0</span>,<span class="built_in"><span class="keyword">sizeof</span></span>(g));</span><br><span class="line">        <span class="built_in">scanf</span>(<span class="string">&quot;%lld%lld%lld%lld&quot;</span>,&amp;n,&amp;m,&amp;k,&amp;mod);</span><br><span class="line">        g[<span class="number">61</span>][<span class="number">1</span>][<span class="number">1</span>][<span class="number">1</span>]=<span class="number">1</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">60</span>;i&gt;=<span class="number">0</span>;i--)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">int</span> x=(n&gt;&gt;i)&amp;<span class="number">1</span>,y=(m&gt;&gt;i)&amp;<span class="number">1</span>,z=(k&gt;&gt;i)&amp;<span class="number">1</span>;</span><br><span class="line">            <span class="keyword">for</span>(<span class="keyword">int</span> a=<span class="number">0</span>;a&lt;<span class="number">2</span>;a++)</span><br><span class="line">             <span class="keyword">for</span>(<span class="keyword">int</span> b=<span class="number">0</span>;b&lt;<span class="number">2</span>;b++)</span><br><span class="line">              <span class="keyword">for</span>(<span class="keyword">int</span> c=<span class="number">0</span>;c&lt;<span class="number">2</span>;c++)</span><br><span class="line">               <span class="keyword">if</span>(g[i+<span class="number">1</span>][a][b][c])</span><br><span class="line">               &#123;</span><br><span class="line">              	 <span class="keyword">for</span>(<span class="keyword">int</span> xx=<span class="number">0</span>;xx&lt;<span class="number">2</span>;xx++)</span><br><span class="line">              	  <span class="keyword">for</span>(<span class="keyword">int</span> yy=<span class="number">0</span>;yy&lt;<span class="number">2</span>;yy++)</span><br><span class="line">              	  &#123;</span><br><span class="line">              	  	<span class="keyword">int</span> zz=xx^yy;</span><br><span class="line">              	   	<span class="keyword">if</span>((a&amp;&amp;x&lt;xx)||(b&amp;&amp;y&lt;yy)||(c&amp;&amp;z&gt;zz)) <span class="keyword">continue</span>;</span><br><span class="line">              	   	<span class="keyword">int</span> aa=(a&amp;&amp;x==xx),bb=(b&amp;&amp;y==yy),cc=(c&amp;&amp;z==zz);</span><br><span class="line">              	   	g[i][aa][bb][cc]=(g[i][aa][bb][cc]+g[i+<span class="number">1</span>][a][b][c])%mod;</span><br><span class="line">              	   	f[i][aa][bb][cc]=(f[i][aa][bb][cc]+f[i+<span class="number">1</span>][a][b][c]+((zz-z)+mod)%mod*((<span class="number">1ll</span>&lt;&lt;i)%mod)%mod*g[i+<span class="number">1</span>][a][b][c]%mod)%mod;</span><br><span class="line">              	  &#125;</span><br><span class="line">               &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="built_in">printf</span>(<span class="string">&quot;%lld\n&quot;</span>,f[<span class="number">0</span>][<span class="number">0</span>][<span class="number">0</span>][<span class="number">0</span>]);</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

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